Solve ∫ (from - 2 to 1) of (2x^2+5x-3)-(3x+1)

$$\int _ { - 2 } ^ { 1 } ( 2 x ^ { 2 } + 5 x - 3 ) - ( 3 x + 1 )$$

Evaluate

$$-9$$

Solution Steps

To find the opposite of 3x+1, find the opposite of each term.

$$\int _{-2}^{1}2x^{2}+5x-3-3x-1\mathrm{d}x$$

Combine 5x and -3x to get 2x.

$$\int _{-2}^{1}2x^{2}+2x-3-1\mathrm{d}x$$

Subtract 1 from -3 to get -4.

$$\int _{-2}^{1}2x^{2}+2x-4\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int 2x^{2}+2x-4\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{2}\mathrm{d}x+\int 2x\mathrm{d}x+\int -4\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{2}\mathrm{d}x+2\int x\mathrm{d}x+\int -4\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 2 times \frac{x^{3}}{3}.

$$\frac{2x^{3}}{3}+2\int x\mathrm{d}x+\int -4\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 2 times \frac{x^{2}}{2}.

$$\frac{2x^{3}}{3}+x^{2}+\int -4\mathrm{d}x$$

Find the integral of -4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{2x^{3}}{3}+x^{2}-4x$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$\frac{2}{3}\times 1^{3}+1^{2}-4-\left(\frac{2}{3}\left(-2\right)^{3}+\left(-2\right)^{2}-4\left(-2\right)\right)$$

Simplify.

$$-9$$